Why Intersect a Basis with a Subspace?
A basis reduces the description of a topology to a selected family of open sets: every open set is a union of basis elements. For a subset \(A\) of a space \(X\), the subspace topology uses the portions of open sets that lie in \(A\). The corresponding basic sets are therefore obtained by intersecting the basis elements of \(X\) with \(A\). This gives a direct way to calculate open sets in a subspace without first listing every open set of the ambient space.
Throughout, let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). A family \(\mathcal{B}\) is a basis for \(\mathcal{T}\) if its members are open, every point of \(X\) belongs to at least one member of \(\mathcal{B}\), and whenever \(x\in B_1\cap B_2\) for \(B_1,B_2\in\mathcal{B}\), there is \(B_3\in\mathcal{B}\) such that \(x\in B_3\subseteq B_1\cap B_2\). Equivalently, every open set is a union of members of \(\mathcal{B}\).
The subspace topology on \(A\) consists of the sets \(A\cap U\) for \(U\in\mathcal{T}\). The intersection is important: an open set of the subspace need not itself be open in \(X\).
Some intersections \(B\cap A\) may be empty. Empty members cause no difficulty: they contribute nothing to unions, and the basis condition involving a point applies only when the intersection contains that point. If \(A=\varnothing\), the subspace has only the open set \(\varnothing\), and the theorem holds as well.
Proof. First, each member \(B\cap A\) of \(\mathcal{B}_A\) is open in the subspace topology, because \(B\in\mathcal{B}\) is open in \(X\). The family covers \(A\): if \(x\in A\), the basis property on \(X\) gives \(B\in\mathcal{B}\) with \(x\in B\), and then \(x\in B\cap A\).
Now let \(O\) be open in the subspace topology. By definition, \(O=A\cap U\) for some \(U\in\mathcal{T}\). Since \(\mathcal{B}\) is a basis, \(U\) is a union of basis elements, say \(U=\bigcup_{B\in\mathcal{C}}B\) for some \(\mathcal{C}\subseteq\mathcal{B}\). Therefore \[ O=A\cap\bigcup_{B\in\mathcal{C}}B =\bigcup_{B\in\mathcal{C}}(A\cap B), \] so every subspace-open set is a union of members of \(\mathcal{B}_A\).
For the basis intersection condition, suppose \(x\in(B_1\cap A)\cap(B_2\cap A)\), where \(B_1,B_2\in\mathcal{B}\). Then \(x\in B_1\cap B_2\). The basis property on \(X\) provides \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\). Since \(x\in A\), we have \(x\in B_3\cap A\), and \(B_3\cap A\subseteq(B_1\cap A)\cap(B_2\cap A)\). Thus the family covers \(A\), consists of subspace-open sets, and satisfies the basis intersection condition. It is a basis for the subspace topology. \(\square\)
Calculating with the Inherited Basis
In the real line with its standard topology, the open intervals form a basis. The theorem says that their intersections with \(A\) form a basis on \(A\). These intersections can have endpoints that are included in the subspace even though the original intervals are open in \(\mathbb{R}\). The interval calculations below illustrate how the ambient interval is clipped by the chosen subset.
Worked Example: A Relative Open Set in a Closed Interval
Let \(A=[0,2]\subseteq\mathbb{R}\), with the subspace topology inherited from the standard topology on \(\mathbb{R}\). The basis elements inherited from the open-interval basis have the form \((p,q)\cap[0,2]\), where \(p<q\).
Consider the ambient open set \(U=(-1,1)\cup(3/2,3)\). Intersecting it with \(A\) gives \[ A\cap U=[0,1)\cup(3/2,2]. \] Indeed, \(x\in[0,2]\cap(-1,1)\) means \(0\le x<1\), giving \([0,1)\); and \(x\in[0,2]\cap(3/2,3)\) means \(3/2<x\le2\), giving \((3/2,2]\).
This set is a union of inherited basis elements: \[ [0,1)=(-1,1)\cap[0,2],\qquad (3/2,2]=(3/2,3)\cap[0,2]. \] The endpoint tests follow directly from the strict inequalities defining the ambient intervals: \(0\) is in \((-1,1)\), and \(2\) is in \((3/2,3)\). Thus the displayed pieces are members of the inherited basis, and their union is open in the subspace.
Check: the first piece includes \(0\) because \(-1<0<1\), but excludes \(1\) because the interval requires \(x<1\). The second excludes \(3/2\) because it requires \(x>3/2\), and includes \(2\) because \(3/2<2<3\).
The inherited basis is a calculation tool, not merely a way to restate the definition. To decide whether a particular subset of \(A\) is open, it is enough to express it as a union of intersections \(B\cap A\). Conversely, the theorem guarantees that every subspace-open set can be expressed in this way.
Worked Example: A Sequence Together with Its Limit
Let \(A=\{0\}\cup\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\), with the subspace topology inherited from the standard topology. Use the open-interval basis on \(\mathbb{R}\). For the interval \((-1/10,3/10)\), compute its intersection with \(A\).
The point \(0\) belongs to the interval because \(-1/10<0<3/10\). Each \(1/n\) is positive, so it automatically satisfies the lower inequality. For the upper inequality, \(1/n<3/10\) is equivalent, since \(n>0\), to \(10<3n\), or \(n>10/3\). For an integer \(n\in\mathbb{N}\), this means \(n\ge4\). The boundary cases confirm the strictness: \(1/3>3/10\), whereas \(1/4<3/10\). Hence \[ A\cap(-1/10,3/10)=\{0\}\cup\{1/n:n\ge4\}. \] This is one member of the inherited basis and is therefore open in \(A\).
For another inherited basis element, use \((1/5,1/2)\). Membership of \(1/n\) requires both \(1/5<1/n\) and \(1/n<1/2\). Multiplying by positive denominators gives \(n<5\) and \(n>2\), respectively, so the only natural-number possibilities are \(n=3\) and \(n=4\). Neither endpoint is included: \(1/5\) corresponds to \(n=5\), and \(1/2\) corresponds to \(n=2\). Therefore \[ A\cap(1/5,1/2)=\{1/3,1/4\}. \] The calculation checks the strict inequalities in both directions and gives a second explicit basis member.
A Finite Example with a Non-Interval Basis
The inherited-basis theorem does not depend on an order or a metric. A basis can be useful even when its members are not intervals, as the following finite example shows.
Worked Example: Inheriting a Basis on a Three-Point Space
Let \(X=\{a,b,c\}\), and let \[ \mathcal{B}=\{\{a\},\{b,c\},X\}. \] Give \(X\) the topology \(\mathcal{T}=\{\varnothing,\{a\},\{b,c\},X\}\). The family \(\mathcal{B}\) is a basis for this topology: its members cover \(X\), and every open set listed in \(\mathcal{T}\) is a union of members of \(\mathcal{B}\). For the intersection condition, intersections with \(X\) leave the other basis member unchanged; \(\{a\}\cap\{b,c\}=\varnothing\) contains no point requiring a refinement; and each basis member intersects itself in itself.
Take the subset \(A=\{a,c\}\). Intersecting each basis element with \(A\) gives \[ \{a\}\cap A=\{a\},\qquad \{b,c\}\cap A=\{c\},\qquad X\cap A=\{a,c\}. \] Thus the inherited basis is \(\mathcal{B}_A=\{\{a\},\{c\},\{a,c\}\}\). Its unions are exactly \(\varnothing,\{a\},\{c\},\{a,c\}\), so these are precisely the open sets of the subspace.
Check: each of the four subsets just listed is a union of inherited basis elements: the empty union gives \(\varnothing\), the individual singleton members give \(\{a\}\) and \(\{c\}\), and their union gives \(A\). There are only four subsets of a two-point set, so no other subspace-open sets are missing.
A Common Mistake
Pitfall: It is incorrect to form a subspace basis by keeping only those ambient basis elements that are already contained in \(A\). For \(A=[0,2]\subseteq\mathbb{R}\), this would keep only open intervals contained in \([0,2]\). No such interval contains \(0\): every open interval containing \(0\) also contains negative real numbers, so it cannot be a subset of \([0,2]\). The proposed family would fail to cover \(A\), and therefore would not be a basis.
The correction is to intersect every ambient basis element with \(A\), not to discard basis elements that extend outside \(A\). For example, \((-1,1)\) is not contained in \([0,2]\), but its intersection with \(A\) is \([0,1)\), an inherited basis element containing \(0\). The theorem's intersection operation preserves the points of \(A\) while removing the portion outside the subspace.
This construction matters because it transfers a known basis to a smaller space with no need to reconstruct its topology from scratch. In familiar spaces, it turns interval or coordinate calculations into local descriptions on subsets; in finite spaces, it converts a short basis list into a complete list of relative opens by taking unions.
Check Your Understanding
Use the inherited-basis description to answer each question. When calculating an intersection, check whether each endpoint is included.
- In \(\mathbb{R}\), let \(A=[-1,1]\). Compute \(A\cap(-2,0)\) and \(A\cap(0,2)\), and identify both as inherited basis elements.
- For \(A=\{0\}\cup\{1/n:n\in\mathbb{N}\}\), compute \(A\cap(1/4,1)\). Which values of \(n\) occur?
- Explain why the family of ambient open intervals contained in \([0,2]\) does not cover \([0,2]\).
- For \(X=\{a,b,c\}\) and \(\mathcal{B}=\{\{a\},\{b,c\},X\}\), list the inherited basis when \(A=\{b,c\}\).
- State why every open set in a subspace \(A\) is a union of sets \(B\cap A\), where \(B\) ranges over a basis for the ambient space.